Harvard-MIT Division of Health Sciences and Technology
HST.584J: Magnetic Resonance Analytic, Biochemical, and Imaging Techniques, Spring 2006
Course Directors: Dr. Bruce Rosen and Dr. Lawrence Wald
HST.584 / 22.561 Exam Solutions
Marking Scheme: a, b, c, d, e, f, h, i, j, k, l, m – 5 marks; g – 20 marks; Total = 70 marks
as students could omit 2 questions.
a) As stated, we assume all magnetization is completely relaxed (i.e. along the zaxis) at the start of the pulse sequence. Following a right-handed rotation of 45°
r M
about the –y axis, we have M = 0 (−1,0,1) .
2
b) Given the constraints, after 99 ms any transverse components will have
*
experienced T2* dephasing of the form e − t T2 . Therefore,
r M
M = 0 (−e −97 50 ,0,1) = M 0 ( −0.102,0, 1 ) .
2
2
c) We desire the same signal amplitude after our second pulse as we had after our
first pulse. This means we require the same transverse component (transverse
component determines signal amplitude) i.e. Mx(97ms) = Mx(2ms). If T1 >> 101
ms (i.e. no longitudinal relaxation has occurred), just prior to the second pulse we
will assume that all existing transverse magnetization has been dephased (T2*
decay) and our longitudinal component is the same as after our initial pulse.
Therefore, to generate the same Mx component following our second pulse, we
need a 90° pulse. In other words:
r
M
i.
M ( 2ms ) = 0 ( −1,0,1)
2
r
M
ii.
M (99ms ) = 0 (0,0,1)
2
r
M
iii.
M (101ms ) = 0 ( −1,0,0) following 90° pulse about –y axis
2
d) If we account for T1 recovery of the longitudinal magnetization, Mz(99ms) will
be larger than Mz(2ms) and thus we will require a smaller flip angle than in the
previous question.
M z (t ) = M 0 + ( M z (0) − M 0 )e − t T1
M
M z (99ms) = M 0 + ( 0
− M 0 )e −97 1001 = 0.889 M 0
2
− M0
We require Mx(101ms) = Mz(99ms)sinθ2 = 0.889M0 sinθ2 = Mx(2ms) =
.
2
Therefore, θ2 = 52.3° about the –y axis.
e) Our bandwidth determines the slice thickness; our frequency offset from the
Larmor frequency determines the slice position along the z-axis.
i. Bandwidth: ∆zγG Z = 0.1cm(4000G −1 s −1 )(1Gcm −1 ) = 400 Hz
ii.
Offset: z off γGZ = 1cm(4000G −1 s −1 )(1Gcm −1 ) = 4000Hz
f) This is simply a Fourier relationship. Sinc RF pulses produce rectangular
excitation profiles. Gaussian RF pulses produce Gaussian excitation profiles,
with an inverse relationship (i.e. A wide Gaussian in time will excite a narrow
Gaussian slice, a narrow Gaussian in time will produce a wide Gaussian slice). I
am not going to attempt to draw Gaussians in Word – see me if you’d like an
example.
t
g) The basic formula for determining k-space position is k (t ) = γ ∫ G (τ )dτ . Our unit
0
of δk corresponds to δk = γGt = 4000G s (1Gcm )(0.5 x10 s ) = 2cm −1 . To
determine our trajectory, we simply integrate our gradient waveforms.
−1 −1
−1
−3
Fig. 1 – k-space trajectory for the gradient waveforms between t = 0 ms and t = 100
ms. The numbers on the actual trajectory corresponds to times in ms.
h) If we read off the trajectory we traced out in the previous question, we produce
the following ordered pairs (in units of δk):
Time (ms)
(kx, ky) (Units of δk)
0
(0, 0)
11
(-5, 15)
14
(-5, 15)
16
(-5, 31)
99
(-5, 31)
110
(15, -5)
i) This large Gy pulse is a spoiler or crusher gradient meant to dephase the residual
transverse magnetization.
j) No, we did not use a rectilinear approach to cover k-space (constant spacing along
k-space lines Æ think EPI or 2D-FT imaging approach).
k) After the second series of gradient manipulations, we have traced out two zig-zag
patterns about the +ky axis and +kx axis. While we can use Fourier symmetry
relations to reduce the k-space coverage for a real object (and most objects are
real), we have not covered enough of k-space yet to employ one of the relations.
At minimum, we need to cover ½ of k-space to reconstruct a real object (and in
practice slightly more, such as 9/16 or 5/8, in order to make appropriate phase
corrections – for more detail on this take HST.587 with Dr. Elfar Adelsteinsson)
l) Resolution = 1 / kmax in both dimensions.
Taking kxmax = 15 δk = 30 cm-1. Therefore, resx = 1/30 cm
Taking kymax = 15 δk = 30 cm-1. Therefore, resy = 1/30 cm
m) kx is sampled most densely near the origin (between 3 & 4 ms) & least densely on
the final leg of the trajectory (between 10 & 11 ms).
FOVx = 1 / ∆kx, FOVy = 1 / ∆ky
If our FOV is based on the least densely sampled region, we cover 9 δk in 16
samples. Therefore, ∆kx = 9 (2 cm-1) / 16 = 9/8 cm-1. Thus FOVx = 8/9 cm.
Similarly, ∆ky = 9 (2 cm-1) / 16 = 9/8 cm-1. Thus FOVy = 8/9 cm.